Loop in a Singly-Linked List

Question: How would you find a loop in a singly-linked list?
Answer: This is a very typical tech interview question that has a very elegant solution that runs in O(n) time and O(1) space. Also referred to as the “Tortoise and Hare Algorithm”,  the solution uses a fast pointer (traversing the list two items at a time) and a slow pointer (traversing the list one item at a time). If there is a loop, then the fast pointer will go around the loop twice as fast as the slow pointer. At one point, these two pointers will point to the same item. Here’s some code to make this clearer.
bool hasLoop( Node *startNode )
{
    Node *slowNode, *fastNode;
    slowNode = fastNode = startNode;
    while( slowNode && fastNode && fastNode->next )
    {
        if( slowNode == fastNode->next || slowNode == fastNode->next->next )
        {
            return true;
        }
        slowNode = slowNode->next;
        fastNode = fastNode->next->next;
    }
    return false;
}

Comments

Post a Comment

Popular posts from this blog

Pick a Random Byte

Question:  You have a stream of bytes from which you can read one byte at a time. You only have enough space to store one byte. After processing those bytes, you have to return a random byte. Note: The probability of picking any one of those bytes should be equal. Answer:  Since we can only store one byte at a time, we have to be able to work with the bytes as they come in. We can’t store everything, count the number of bytes and pick one. That would be too easy wouldn’t it? Let’s think it through. We don’t know how many bytes there are, so at any point in time we should have a random byte picked from all the bytes we have seen so far. We shouldn’t worry about the total number of bytes at any time. When we get the first byte, it is simple. There is only one so we store it. When we get the second one, it has 1/2 probability of being picked and the one we have stored has a 1/2 probability of being picked (we can use any programming language’s built-in random function). After pick
Read more

The Fox and The Duck

Question:  A duck that is being chased by a fox saves itself by sitting at the center of circular pond of radius r. The duck can fly from land but cannot fly from the water. Furthermore, the fox cannot swim. The fox is four times faster than the duck. Assuming that the duck and fox are perfectly smart, is it possible for the duck to ever reach the edge of the pond and fly away to its escape from the ground? Answer:  One would think that the duck could swim directly away from the fox. So the duck would have to swim a distance r. The fox would have to cover half the circumference of the pond (pi * r). Since the fox is four times faster than the duck, we know that (pi * r) < (4 * r) This would make it seem like it is impossible for the duck to escape. So how could the duck make life most difficult for the fox? If the duck just tries to swim along a radius, the fox could just sit along that radius and the duck would continue to be trapped. The duck could swim in concentric c
Read more

First Non-Repeated Character

Question:  Write an algorithm to find the first non-repeated character in a string. For example, the first non-repeated character in the string ‘abcdab’ is ‘c’. Answer:  Seems trivial enough right? If a character is repeated, we should be able to search the string to determine if that character appears again. So if we go about doing this for every character in the string, our worst case run time would O(n^2). Here is some code to show this logic. char returnFirstNonRepeatedChar( char * str ) { int i, repeated = 0; int len = strlen(str); for( i = 0; i < len; i++ ) { repeated = 0; for( j = 0; j < len; j++ ) { if( i != j && str[i] == str[j] ) { repeated = 1; break; } } if( repeated == 0 ) { // Found the first non-repeated character return str[i]; } } return ''; } This approach seems to work perfectly well. But
Read more